Let $\mathcal{B}$ be the entire Borel set of $\mathbb{R}$. And let $\chi_E$ be the indicator function of the set $E$.

$(X, m, \mu)$; measure space, map $f: X \to \mathbb{R}$; measurable.

For $A \in \mathcal{B}$, we define $$ \lambda(A) = \mu(f^{-1}(A)). $$

When $g = \chi_E$, $E \in \mathcal{B}$, the following equation transformation holds: $$ \int_{\mathbb{R}} \chi_E\ d\lambda = \lambda(E) = \mu(f^{-1}(E)) \overset{?}{=} \int_{X} \chi_E \circ f\ d\mu. $$

Why is the third equal sign valid?

What I know

$\chi_E \circ f = \chi_E(f)$,

$\int_{X} \chi_E \circ f\ d\mu = \mu((\chi_E \circ f) \cap X)$.


1 Answer 1


$\chi_E \circ f$ is not exactly $\chi_E(f)$, as the input set of $\chi_E$ is $\mathbb{R}$. Moreover, your formula $\int_{X} \chi_E \circ f\ d\mu = \mu((\chi_E \circ f) \cap X)$ doesn't make sense as $(\chi_E \circ f) \cap X$ is not a set as $(\chi_E \circ f)$ is a function and not a set.

More precisely, $\chi_E \circ f$ is a function from $\mathbb{R}$ to $\mathbb{R}$ defined by

$$\forall x \in \mathbb{R}, \quad (\chi_E \circ f) (x) = \chi_E(f(x)).$$

Therefore we can see that $(\chi_E \circ f)= \chi_{f^{-1}(E)}$. Indeed, for $x \in \mathbb{R}$, $$\begin{array}{lll} (\chi_E \circ f) (x) &=& \chi_E(f(x)) \\ &=& \left\{\begin{array}{lll}1 &\text{si }f(x) \in E \\ 0 &\text{otherwise} \end{array} \right. \\ &=& \left\{\begin{array}{lll}1 &\text{si }x \in f^{-1}(E) \\ 0 &\text{otherwise} \end{array} \right. \\ &=& \chi_{f^{-1}(E)}(x) \end{array}$$

Therefore we get $$\mu(f^{-1}(E)) = \displaystyle\int_X \chi_{f^{-1}(E)} d\mu = \displaystyle\int_X \chi_E \circ f d\mu.$$

  • $\begingroup$ Thank you for letting us know about the mistake! I understood. $\endgroup$
    – ytnb
    Dec 1, 2022 at 9:16

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